Mathematics Grade 12 Paper 1 Challenge Paper 2026 Memo

Copyright reserved Please turn over hlayiso mock exams Practice Series: Grade 12 NSC Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P1 CHALLENGE PAPER 2026 MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 11 pages.

Mathematics/P1 2 NSC – Marking Guidelines Hlayiso/Challenge 2026 NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • If a candidate has crossed out an attempt of a question and not redone the question, mark the crossed-out version. • Consistent accuracy (CA) applies in ALL aspects of the marking guidelines. Stop marking at the second calculation error. • Assuming answers/values in order to solve a problem is NOT acceptable. In show that questions the given answer earns no mark: marks are for the working that leads to it. • Valid alternative methods that are not shown here must be marked on merit, using the same mark allocation. QUESTION 1 1.1.1 x4 +12 = 7x2 (x̸ = 0) x4 −7x2 +12 = 0 (x2 −3)(x2 −4) = 0 x = ± √ 3 or x = ±2 ✓multiply by x2 ✓standard form ✓factors ✓all four answers (answers only: max 2/4) (4) 1.1.2 3·3x + 3 3x = 10 Let k = 3x: 3k2 −10k +3 = 0 (3k −1)(k −3) = 0 3x = 1 3 ⇒x = −1 or 3x = 3 ⇒x = 1 ✓3·3x and 3·3−x ✓standard form ✓factors ✓x = ±1 (4) 1.2 12+x−x2 ≥0 ⇒x2 −x−12 ≤0 (x−4)(x+3) ≤0 ⇒−3 ≤x ≤4 and x̸ = 2 −3 ≤x ≤4, x̸ = 2 (OR x ∈[−3 ; 2)∪(2 ; 4]) ✓12+x−x2 ≥0 ✓−3 ≤x ≤4 ✓x̸ = 2 (3) 1.3 2x ·22y = 25 ⇒x+2y = 5 ⇒x = 5−2y (5−2y)2 +2y2 = 11 6y2 −20y+14 = 0 ⇒3y2 −10y+7 = 0 (3y−7)(y−1) = 0 y = 1 and x = 3 or y = 7 3 and x = 1 3 ✓x+2y = 5 ✓substitution ✓standard form ✓y-values ✓x-values (5) 1.4.1 ∆= [−2(k +1)]2 −4(k −1)(k +3) = 4(k2 +2k +1)−4(k2 +2k −3) = 16 = 42 ∆is a perfect square (independent of k) and the coefficients are rational, so the roots are rational for every rational k̸ = 1. ✓substitution into ∆ ✓∆= 16 ✓conclusion: perfect square (3) Copyright reserved Please turn over

Mathematics/P1 3 NSC – Marking Guidelines Hlayiso/Challenge 2026 1.4.2 x = 2(k +1)±4 2(k −1) x = 1 or x = k +3 k −1 = 1+ 4 k −1 k −1 must be a factor of 4: k −1 ∈{±1 ; ±2 ; ±4} k ∈{−3 ; −1 ; 0 ; 2 ; 3 ; 5} ✓roots 1 and k+3 k−1 ✓1+ 4 k−1 / factors of 4 ✓all six values (3) 1.5 xy = ( √ 3)2 −( √ 2)2 = 1 x2 +y2 = (5+2 √ 6)+(5−2 √ 6) = 10 x y + y x = x2 +y2 xy = 10 1 = 10 ✓xy = 1 ✓x2 +y2 = 10 ✓answer 10 (3) [25] QUESTION 2 2.1.1 First differences: p+7 ; 5−p ; q−5 ; 33−q Second differences equal: (5−p)−(p+7) = (q−5)−(5−p) ⇒q = 8−3p (q−5)−(5−p) = (33−q)−(q−5) ⇒p+3q = 48 p+3(8−3p) = 48 ⇒p = −3; q = 8+9 = 17 OR: Tn = an2 +bn+c with T1 = −7, T3 = 5, T5 = 33: 8a+2b = 12; 16a+2b = 28 ⇒a = 2, b = −2, c = −7 p = T2 = −3; q = T4 = 17 ✓first differences ✓first equation ✓second equation ✓solving OR: ✓3 equations ✓✓a, b, c ✓T2 and T4 (4) 2.1.2 Second difference: (5−p)−(p+7) = 8−4 = 4 ⇒2a = 4, a = 2 3a+b = T2 −T1 = 4 ⇒b = −2 a+b+c = −7 ⇒c = −7 Tn = 2n2 −2n−7 ✓a = 2 ✓b = −2 ✓c = −7 (3) 2.1.3 Tn = 2n(n−1)−7 2n(n−1) is even, so Tn = even −odd = odd for every n. 2 026 is even, so it is not a term. OR: 2n2 −2n−7 = 2026 ⇒2n2 −2n−2033 = 0 n = 2± √ 16268 4 = 32,39 or −31,39; n /∈N ✓2n(n−1) even / equation ✓conclusion: Tn odd / n /∈N (2) Copyright reserved Please turn over

Mathematics/P1 4 NSC – Marking Guidelines Hlayiso/Challenge 2026 2.2 log2 (8r 2 ) = log2 ( 23r−1) = 3r −1 k ∑ r=1 (3r −1) = 2+5+8+... (a = 2, d = 3) k 2 [2(2)+(k −1)(3)] = 100 3k2 +k −200 = 0 (3k +25)(k −8) = 0 k = 8 (k̸ = −25 3 ) ✓3r −1 ✓a = 2 and d = 3 ✓substitution into Sn = 100 ✓k = 8 (with rejection) (4) [13] QUESTION 3 3.1.1 T1 = a; T3 = a+2d; T7 = a+6d a+2d a = a+6d a+2d ⇒(a+2d)2 = a(a+6d) a2 +4ad +4d2 = a2 +6ad 4d2 = 2ad ⇒2d(2d −a) = 0 d̸ = 0 ∴a = 2d ✓a+2d and a+6d ✓equal ratios ✓simplification with d̸ = 0 (3) 3.1.2 S10 = 10 2 [2(2d)+9d] = 65d = 260 d = 4; a = 8 Geometric: 8 ; 16 ; 32 ; ... r = 2 T5 = 8(2)4 = 128 ✓65d = 260 ✓a = 8, d = 4 and r = 2 ✓128 (3) 3.2.1 r = 1−2x −1 < 1−2x < 1 −2 < −2x < 0 ⇒0 < x < 1 ✓−1 < 1−2x < 1 ✓0 < x < 1 (2) 3.2.2 12 1−r = 20 ⇒r = 0,4 S∞−Sn = 20−12(1−0,4n) 1−0,4 = 20(0,4)n 20(0,4)n < 0,01 ⇒(0,4)n < 0,0005 n > log0,0005 log0,4 = 8,30 (inequality reverses) At least 9 terms (Note: x = 0,3, but x is not needed.) ✓r = 0,4 ✓20(0,4)n / S∞−Sn ✓use of logs ✓9 terms (4) [12] Copyright reserved Please turn over

Mathematics/P1 5 NSC – Marking Guidelines Hlayiso/Challenge 2026 QUESTION 4 4.1 f(x) = 2(x−1)+3 x−1 = 2(x−1) x−1 + 3 x−1 = 2+ 3 x−1 OR: 3 x−1 +2 = 3+2(x−1) x−1 = 2x+1 x−1 ✓2x+1 = 2(x−1)+3 / common denominator ✓splitting the fraction (2) 4.2 x = 1 y = 2 ✓x = 1 ✓y = 2 (2) 4.3 mx = 2x+1 x−1 ⇒mx2 −mx = 2x+1 mx2 −(m+2)x−1 = 0 No intersection: ∆< 0 (m+2)2 +4m < 0 ⇒m2 +8m+4 < 0 Critical values: m = −8± √ 48 2 = −4±2 √ 3 −4−2 √ 3 < m < −4+2 √ 3 (−7,46 < m < −0,54) (For m = 0 the line y = 0 cuts f at x = −1 2; x = 1 is never a root.) ✓standard form ✓∆< 0 ✓m2 +8m+4 < 0 ✓answer (4) 4.4 g(x) = −f(−x): reflection in the y-axis followed by reflection in the x-axis (i.e. rotation through 180◦about the origin) OR: translation 2 units to the left and 4 units down (The two reflections may be in either order.) ✓✓correct description (OR ✓2 left ✓4 down) (2) [10] QUESTION 5 5.1 O(0 ; 0): 2p +q = 0 …(1) B(2 ; 12): 2p+2 +q = 12 ⇒4·2p +q = 12 …(2) (2)−(1): 3·2p = 12 ⇒2p = 4 ⇒p = 2 q = −2p = −4 ✓equation (1) ✓equation (2) ✓2p = 4 ✓p = 2 and q = −4 (4) 5.2 h : y = 2x+2 −4 h−1 : x = 2y+2 −4 2y+2 = x+4 ⇒y+2 = log2(x+4) y = log2(x+4)−2 ✓swop x and y ✓log2(x+4) ✓answer (3) 5.3 x > −4 (OR x ∈(−4 ; ∞)) ✓answer (1) Copyright reserved Please turn over

Mathematics/P1 6 NSC – Marking Guidelines Hlayiso/Challenge 2026 5.4 log2(x+4)−2 ≤2 ⇒log2(x+4) ≤4 x+4 ≤16 ⇒x ≤12 Domain: x > −4 ∴−4 < x ≤12 (OR: B(2;12) on h ⇒(12;2) on h−1, and h−1 is increasing.) ✓x+4 ≤16 / point (12;2) ✓x ≤12 ✓−4 < x (3) [11] QUESTION 6 6.1 Axis of symmetry: x = −1+5 2 = 2 D on y = 3x−15: y = 3(2)−15 = −9; D(2 ; −9) f(2) = a(3)(−3) = −9a = −9 ⇒a = 1 ✓x = 2 ✓y = −9 ✓substitution: −9a = −9(3) 6.2 f(x) = x2 −4x−5 f ′(x) = 2x−4 = 2 ⇒x = 3 f(3) = 9−12−5 = −8; −8 = 2(3)+k ⇒k = −14 OR: x2 −4x−5 = 2x+k ⇒x2 −6x−(5+k) = 0 ∆= 36+4(5+k) = 0 ⇒k = −14 ✓f ′(x) = 2 ✓point (3 ; −8) ✓k = −14 OR: ✓equating ✓∆= 0 ✓ k (3) 6.3 Two distinct points: k > −14 (line above the tangent) The line y = 2x+k passes through (0 ; −5), the y-intercept of f, when k = −5; then the points of intersection have x = 0 and x = 6. For both x-values to be positive the line must lie below this position: k < −5 −14 < k < −5 OR: x2 −6x−(5+k) = 0 ⇒x = 3± √ 14+k Distinct: 14+k > 0; smaller root positive: √ 14+k < 3 ⇒k < −5 ✓k > −14 ✓k < −5 (with reason) ✓answer (3) 6.4 x-intercepts of g: x = −1+t and x = 5+t Both positive: −1+t > 0 ⇒t > 1 ✓−1+t ✓t > 1 (2) 6.5 f(x) > 0 for x < −1 or x > 5; f(x) < 0 for −1 < x < 5 f ′(x) = 2x−4: f ′(x) > 0 for x > 2; f ′(x) < 0 for x < 2 Same sign: −1 < x < 2 or x > 5 ✓critical values −1; 2; 5 ✓−1 < x < 2 ✓x > 5 (3) [14] Copyright reserved Please turn over

Mathematics/P1 7 NSC – Marking Guidelines Hlayiso/Challenge 2026 QUESTION 7 7.1 After 3 years: A = 50000 ( 1+ 0,08 4 )12 = R63412,09 100000 = 63412,09 ( 1+ 0,09 12 )m (1,0075)m = 1,57698... m = log1,57698... log1,0075 = 60,96 61 months after the rate change Total: 36+61 = 97 months ✓i = 0,02 and n = 12 ✓R63 412,09 ✓equation with i = 0,0075 ✓m = 60,96 ✓97 months (5) 7.2.1 i = 0,12 12 = 0,01 Balance after 2 months: 420000(1,01)2 = R428442 428442 = x [ 1−(1,01)−72] 0,01 x = R8376,12 OR: 420000 = x [ 1−(1,01)−72] 0,01 (1,01)−2 ✓i = 0,01 and n = 72 ✓420000(1,01)2 ✓substitution into Pv ✓x = 8376,12 (must be shown) (4) 7.2.2 Remaining instalments: 72−30 = 42 Balance = 8376,12 [ 1−(1,01)−42] 0,01 = R286112,41 (Accept the future-value method: 428442(1,01)30 −8376,12 [ (1,01)30 −1 ] 0,01 = R286112,66.) New balance: 286112,41−60000 = R226112,41 226112,41 = 8376,12[1−(1,01)−n] 0,01 (1,01)−n = 0,73005... ⇒n = −log0,73005... log1,01 = 31,62 She will pay 31 more full instalments (and a smaller final payment). ✓n = 42 ✓substitution ✓R286 112,41 ✓subtract R60 000 ✓use of logs: n = 31,62 ✓31 instalments (6) [15] Copyright reserved Please turn over

Mathematics/P1 8 NSC – Marking Guidelines Hlayiso/Challenge 2026 QUESTION 8 8.1 f(x+h) = 2(x+h)2 − 1 x+h = 2x2 +4xh+2h2 − 1 x+h f(x+h)−f(x) = 4xh+2h2 − 1 x+h + 1 x = 4xh+2h2 + h x(x+h) f ′(x) = lim h→0 4xh+2h2 + h x(x+h) h = lim h→0 ( 4x+2h+ 1 x(x+h) ) = 4x+ 1 x2 ✓f(x+h) ✓ h x(x+h) ✓formula with limit ✓dividing by h ✓answer (Incorrect notation: −1) (5) 8.2 (x+3)2 −9 = x2 +6x y = x2 +6x x 1 3 = x 5 3 +6x 2 3 dy dx = 5 3x 2 3 +4x−1 3 ✓x2 +6x ✓x 5 3 +6x 2 3 ✓answer (3) 8.3 dy dx = 3x2 −3; mtangent = 3p2 −3 Tangent: y−(p3 −3p) = (3p2 −3)(x−p) Through (0 ; 2): 2−p3 +3p = −3p3 +3p 2p3 = −2 ⇒p3 = −1 ⇒p = −1 P(−1 ; 2) (Check: tangent at P is y = 2, which passes through (0;2).) OR: m = p3 −3p−2 p−0 = 3p2 −3 ⇒p3 −3p−2 = 3p3 −3p ✓3p2 −3 ✓substitution of (0;2) into tangent / gradient equation ✓p3 = −1 ✓P(−1 ; 2) (4) [12] QUESTION 9 9.1 f ′(x) = 3x2 +2px+q f ′(2) = 0: 12+4p+q = 0 ⇒4p+q = −12 …(1) f(2) = 0: 8+4p+2q+16 = 0 ⇒2p+q = −12 …(2) (1)−(2): 2p = 0 ⇒p = 0; q = −12 OR: f(x) = (x−2)2(x−c); f(0) = −4c = 16 ⇒c = −4 f(x) = (x−2)2(x+4) = x3 −12x+16 ✓f ′(x) ✓f ′(2) = 0 ✓f(2) = 0 ✓solving simultaneously OR: ✓form ✓c = −4 ✓✓expansion and comparing (4) Copyright reserved Please turn over

Mathematics/P1 9 NSC – Marking Guidelines Hlayiso/Challenge 2026 9.2 f(x) = x3 −12x+16 f ′(x) = 3x2 −12 = 0 ⇒x = ±2 f(−2) = −8+24+16 = 32; f(2) = 0 (−2 ; 32) and (2 ; 0) ✓f ′(x) = 0 ✓x = ±2 ✓y-values (3) 9.3 x y −4 2 16 (−2;32) f f(x) = (x−2)2(x+4): x-intercepts −4 and 2 ✓x-intercepts −4 and 2 and y-intercept 16 ✓turning points ✓shape (3) 9.4 f(x) = k: horizontal line y = k cuts f. For 0 < k < 16 the line cuts f once for x < −2, once for 0 < x < 2 and once for x > 2. For k = 16 one root is x = 0 (not positive); for k = 0 the positive root is repeated. 0 < k < 16 ✓k > 0 ✓k < 16 (reference to y-intercept) ✓strict inequalities (3) 9.5 f is decreasing for −2 < x < 2 g is f shifted 1 unit right: −1 < x < 3 (Accept −1 ≤x ≤3.) ✓−2 < x < 2 / g′(x) = 3(x−1)2 −12 ✓−1 < x < 3 (2) [15] Copyright reserved Please turn over

Mathematics/P1 10 NSC – Marking Guidelines Hlayiso/Challenge 2026 QUESTION 10 10.1 πr2h = 500π ⇒h = 500 r2 C = 0,04 ( 2πr2) +0,02(2πrh) = 0,08πr2 +0,04πr · 500 r2 = 0,08πr2 + 20π r ✓h = 500 r2 ✓cost of top/bottom and side ✓substitution of h (3) 10.2 C′(r) = 0,16πr −20π r2 = 0 0,16r3 = 20 ⇒r3 = 125 ⇒r = 5 cm (C′′(r) = 0,16π + 40π r3 > 0, so this is a minimum.) C(5) = 0,08π(25)+ 20π 5 = 2π +4π = 6π Minimum cost = R18,85 ✓C′(r) ✓C′(r) = 0 ✓r = 5 ✓substitution r = 5 ✓R18,85 (5) [8] QUESTION 11 11.1.1 Independent: P(A and B) = x(x+0,2) 0,76 = x+(x+0,2)−x(x+0,2) x2 −1,8x+0,56 = 0 ⇒(x−0,4)(x−1,4) = 0 (OR 25x2 −45x+14 = 0 ⇒(5x−2)(5x−7) = 0) x = 0,4 (x = 1,4 > 1, rejected) ✓P(A and B) = x(x+0,2) ✓substitution into the addition rule ✓standard form ✓x = 0,4 with rejection (4) 11.1.2 P(A and B) = 0,4×0,6 = 0,24 P(exactly one) = 0,76−0,24 = 0,52 OR: 0,4(0,4)+0,6(0,6) = 0,52 ✓0,24 / method ✓0,52 (2) 11.2.1 Treat the 3 girls as one unit: 5 units, 5! arrangements Girls among themselves: 3! 5!×3! = 720 ✓5! ✓3! ✓720 (3) Copyright reserved Please turn over

Mathematics/P1 11 NSC – Marking Guidelines Hlayiso/Challenge 2026 11.2.2 Seat the 4 boys: 4! = 24 This creates 5 gaps (including the two ends); girls in 3 different gaps: 5×4×3 = 60 n(E) = 24×60 = 1440; n(S) = 7! = 5040 P = 1440 5040 = 2 7 ≈0,29 OR: girls’ seat sets with no two adjacent: {1,3,5},{1,3,6},{1,3,7},{1,4,6},{1,4,7}, {1,5,7},{2,4,6},{2,4,7},{2,5,7},{3,5,7}: 10 sets; 10×3!×4! = 1440 ✓4! ✓5×4×3 ✓7! ✓2 7 (4) 11.2.3 Together: 2×6! = 1440 P(not together) = 1−1440 5040 = 5 7 ≈0,71 ✓2×6! ✓5 7 (2) [15] TOTAL: 150 Copyright reserved